Thus
  \begin{equation*}%
    P\{A_1 \cup A_2\} =
    \frac{1}{2}+\frac{1}{2}-\frac{1}{4}=\frac{3}{4}.
  \end{equation*}

The probability $P\{A_1 \cup A_2\cup \cdots A_n\}$ of the realization of at
  least one among $n$ events can be computed by a formula analogous to (7.4),
  derived in IV,1.
Here we note only that the argument leading to (7.3) applies to any number of
  terms.
Thus \emph{for arbitrary events $A_1, A_2, \ldots$ the inequality}
  \begin{equation}
    P\{A_1 \cup A_2\cup \cdots \} \leq P\{A_1}+P\{A_2}+\ldots
  \end{equation}
  holds.
